Theory Of Equations Ques 27
- If $m$ is chosen in the quadratic equation $\left(m^{2}+1\right) x^{2}-3 x+\left(m^{2}+1\right)^{2}=0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is
(a) $10 \sqrt{5}$
(b) $8 \sqrt{5}$
(c) $8 \sqrt{3}$
(d) $4 \sqrt{3}$
(2019 Main, 9 April II)
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Answer:
Correct Answer: 27.(b)
Solution:
- Given quadratic equation is
$ \left(m^{2}+1\right) x^{2}-3 x+\left(m^{2}+1\right)^{2}=0 ……(i) $
Let the roots of quadratic Eq. (i) are $\alpha$ and $\beta$, so $\alpha+\beta=\frac{3}{m^{2}+1}$ and $\alpha \beta=m^{2}+1$
According to the question, the sum of roots is greatest and it is possible only when " $\left(m^{2}+1\right)$ is minimum" and “minimum value of $m^{2}+1=1$, when $m=0$ “.
$\therefore \alpha+\beta=3$ and $\alpha \beta=1$, as $m=0$
Now, the absolute difference of the cubes of roots
$ \begin{aligned} & =\left|\alpha^{3}-\beta^{3}\right| \\ & =|\alpha-\beta|\left|\alpha^{2}+\beta^{2}+\alpha \beta\right| \\ & =\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}\left|(\alpha+\beta)^{2}-\alpha \beta\right| \\ & =\sqrt{9-4}|9-1|=8 \sqrt{5} \end{aligned} $
BETA
